2002 AMC 8 第 25 题

先试着解答 2002 AMC 8 第 25 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2002 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

25.

Loki、Moe、Nick 和 Ott 是好朋友。Ott 没有钱,但其他人有钱。Moe 给了 Ott 自己钱数的五分之一,Loki 给了 Ott 自己钱数的四分之一,Nick 给了 Ott 自己钱数的三分之一。每个人给 Ott 的钱数相同。现在 Ott 拥有这群人总钱数的几分之几?
Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group’s money does Ott now have?
110\dfrac{1}{10}
14\dfrac{1}{4}
13\dfrac{1}{3}
25\dfrac{2}{5}
12\dfrac{1}{2}
答案:B
知识点:分数逆推法
小提示:
假设每个朋友都给 Ott 相同且便于计算的金额。
Assume each friend gives Ott the same convenient amount.
大提示:
如果每人给 Ott $1\$1,反推出每个朋友原来的钱数。
If each gives Ott $1,\$1, work backward to each friend’s original amount.
解答:
因为只有各人所给钱数占原有钱数的比例重要,所以假设每个人都给 Ott $1\$1。这意味着 Moe 原来有 $5\$5,Loki 原来有 $4\$4,Nick 原来有 $3\$3。
转账不会改变这群人的总钱数,即 $5+$4+$3=$12。 \$5 + \$4 + \$3 = \$12\text{。}Ott 现在有 $3\$3。
因此 Ott 拥有这群人总钱数的 312=14, \dfrac{3}{12} = \dfrac{1}{4}\text{,}也就是四分之一。
所以正确答案是 B。
Because only the fractions matter, suppose each person gave Ott $1.\$1. This means that Moe had $5,\$5, Loki had $4,\$4, and Nick had $3\$3 originally.
The transfers do not change the group’s total amount of money, which is $5+$4+$3=$12. \$5 + \$4 + \$3 = \$12. Ott now has $3.\$3.
This means that Ott has 312=14 \dfrac{3}{12} = \dfrac{1}{4} of the group’s money.
Thus, B is the correct answer.
LIVE by Po-Shen Loh
LIVE项目

LIVE直播课程

Self-Paced录播课程

启明星Star计划

奖学金项目

关于我们

罗博深是谁?

邮件联系

订阅英文简报

加入我们

申请明星导师

For a more Thought Full world.

© 2026 Expii, Inc.